Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $r = \dfrac{9z^2 + 27z}{z^2 + 11z + 30} \times \dfrac{-8z - 48}{z + 3} $
Answer: First factor the quadratic. $r = \dfrac{9z^2 + 27z}{(z + 6)(z + 5)} \times \dfrac{-8z - 48}{z + 3} $ Then factor out any other terms. $r = \dfrac{9z(z + 3)}{(z + 6)(z + 5)} \times \dfrac{-8(z + 6)}{z + 3} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ 9z(z + 3) \times -8(z + 6) } { (z + 6)(z + 5) \times (z + 3) } $ $r = \dfrac{ -72z(z + 3)(z + 6)}{ (z + 6)(z + 5)(z + 3)} $ Notice that $(z + 3)$ and $(z + 6)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ -72z(z + 3)\cancel{(z + 6)}}{ \cancel{(z + 6)}(z + 5)(z + 3)} $ We are dividing by $z + 6$ , so $z + 6 \neq 0$ Therefore, $z \neq -6$ $r = \dfrac{ -72z\cancel{(z + 3)}\cancel{(z + 6)}}{ \cancel{(z + 6)}(z + 5)\cancel{(z + 3)}} $ We are dividing by $z + 3$ , so $z + 3 \neq 0$ Therefore, $z \neq -3$ $r = \dfrac{-72z}{z + 5} ; \space z \neq -6 ; \space z \neq -3 $